Calculating interest and duration

Exponential and logarithmic growth: Future value

Calculating interest and duration

Earlier we saw how to calculate the future value and present value. Using the same formula, \(S_n = S_0 \cdot (1+i)^n\), we can also calculate the interest and duration.

Interest on deposit without additional contribution

In the case where an amount is deposited in a bank account and accrues compound interest, and the start value #S_0#, the terminal value #S_n#, and the duration #n# are known, the interest rate can be calculated by using the formula

\[i = \left(\dfrac{S_n}{S_0}\right)^{\frac{1}{n}}-1\]

By applying the same algebraic operations to both sides of the equation, we can rewrite the formula #S_n = S_0 \cdot (1+i)^n# in such a way that #i# only appears at the left of the equality sign.

\[\begin{array}{rcl}

S_0 \cdot (1+i)^n &=&S_n\\

&& \phantom{xxxxx}\color{blue}{\text{the known formula with the two sides interchanged}}\\

(1+i)^n &=& \dfrac{S_n}{S_0} \\

&& \phantom{xxxxx}\color{blue}{\text{both sides divide by } S_0}\\

1+i &=& \left(\dfrac{S_n}{S_0}\right)^{\frac{1}{n}} \\

&& \phantom{xxxxx}\color{blue}{\text{both sides raised to the power } \frac{1}{n}}\\

i &=& \left(\dfrac{S_n}{S_0}\right)^{\frac{1}{n}}-1 \\

&& \phantom{xxxxx}\color{blue}{ 1\text{ subtracted from both sides}}

\end{array}\]

Once we have found the growth rate #i#, we can multiply it by #100# in order to obtain the interest rate.

After filling in values for #S_n#, #S_0#, and #n#, we find #i#.

Similarly, we can also calculate the unknown duration.

Duration with no additional contribution

If an amount is deposited in a bank account and accrues compound interest, and the start value #S_0#, the terminal value #S_n#, and growth rate #i# are known, we can calculate the duration #n# by using the formula

\[n = \log_{1+i}\left({S_n}\right)- \log_{1+i}\left({S_0}\right)\]

By using the same algebraic operations on both sides of the equation, we can rewrite the formula #S_n = S_0 \cdot (1+i)^n# in such a way that the left side of the equality sign becomes equal to #n#.

\[\begin{array}{rcl}
S_n &=& S_0 \cdot (1+i)^n \\
&& \phantom{xxxxx}\color{blue}{\text{the known formula}}\\
(1+i)^n &=& \dfrac{S_n}{S_0} \\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by } S_0}\\
\log_{1+i}((1+i)^n) &=& \log_{1+i}\left(\frac{S_n}{S_0}\right) \\
&& \phantom{xxxxx}\color{blue}{\log_{1+i} \text{ taken at both sides}}\\
n &=& \log_{1+i}\left(\dfrac{S_n}{S_0}\right) \\
&& \phantom{xxxxx}\color{blue}{\text{rule } \log_g(g^n)=n}\\
n &=& \log_{1+i}\left({S_n}\right)- \log_{1+i}\left({S_0}\right) \\
&& \phantom{xxxxx}\color{blue}{\text{rule } \log_g(\frac{x}{y})=\log_g(x)-\log_g(y)}
\end{array}\]

After filling in values for #S_n#, #S_0#, and #i#, we find #n#.

In fact, you only need to remember the formula #S_n=S_0 \cdot (1+i)^n#. As indicated above, you can solve for the unknown variable from it.

In the examples below you can see how this works in practice.

A capital of #\euro \, 10000# matured after a deposit of #\euro \, 3427.29# at the bank #20# years before with a fixed annual interest.

What was the annual interest? Give your answer with a precision of #1# decimal place.

Annual interest rate: #5.5# #\%#

In order to see this, we apply the formula
\[S_n=S_0\cdot (1+i)^n\]
Here #S_n=10000#, #S_0=3427.29#, and #n=20#, so the growth rate #i# satisfies
\[10000=3427.29\cdot(1+i)^{20}\]
We solve this equation for #i# as follows:
\[\begin{array}{rcl}
10000&=&3427.29\cdot(1+i)^{20}\\
&& \phantom{xxxxx}\color{blue}{\text{the equation to be solved}}\\
(1+i)^{20} &=& \dfrac{10000}{3427.29} \\
&& \phantom{xxxxx}\color{blue}{\text{both sides divided by } 3427.29}\\
1+i&=&\left(\dfrac{10000}{3427.29}\right)^{\frac{1}{20}}\\
&& \phantom{xxxxx}\color{blue}{\text{both sides raised to the power } \frac{1}{20}}\\
1+i &\approx& 1.055\\
&& \phantom{xxxxx}\color{blue}{\text{calculated }}\\
i&=&0.055\\
&& \phantom{xxxxx}\color{blue}{1 \text{ subtracted from both sides}}\\
\end{array}\]

Thus, the annual interest rate is #0.055 \cdot 100 = 5.5 \%#.

New example